Conservation of Momentum EX 5

The mass of the third piece is (6 - 3 - 2)kg = 1 kg. We will compute the components of the linear momentum of the bomb, $p_{b}$, the 1-kg piece, $p_{1}$, the 2-kg piece, $p_{2}$, and the 3-kg piece, $p_{3}$. The linear momentum is conserved. We predict that the vector $p_{3}$ must be in the 4th quadrant so that $p_{b}=p_{1}+p_{2}+p_{3}$.

Momentum	x-component	y-component
$p_{b}$	$(6)(5)\cos(-37^{\circ })=24{\tfrac {kgm}{s}}$	$(6)(5)\sin(-37^{\circ })=18{\tfrac {kgm}{s}}$
$p_{1}$	${p_{1}}_{x}$	${p_{1}}_{y}$
$p_{2}$	$-(2)(3)=-6{\tfrac {kgm}{s}}$	0
$p_{3}$	$(3)(2)\cos(53^{\circ })=3.6{\tfrac {kgm}{s}}$	$(3)(2)\sin(53^{\circ })=4.8{\tfrac {kgm}{s}}$

Then we write the equations

$p_{{before}_{x}}=p_{{after}_{x}}$

$24={p_{1}}_{x}-6+3.6$

${p_{1}}_{x}=26.4{\tfrac {kgm}{s}}$

and

$p_{{before}_{y}}=p_{{after}_{y}}$

$-18={p_{1}}_{y}+0+4.8$

${p_{1}}_{y}=-22.8{\tfrac {kgm}{s}}$

To obtain the velocity of the third fragment, we divide its momentum by its mass (1 kg):

${\vec {v}}=26.4{\hat {i}}-22.8{\hat {j}}$

We note that our prediction was right. The linear momentum of the third piece is in the fourth quadrant.